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Help Needed. Proof upper bound of a set.

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A be a set of real numbers. If b is the supremum (least upper bound) of the set A then whenever c<b there exist an a in A such that a>c.


    2. Relevant equations



    3. The attempt at a solution

    I considered two cases. The first one when the supremum b is attained by the set A. In this case there exists an a belonging to A such that a=b and the statement is proved.

    In the second case the supremum is not attained by the set A, so for all a that belong to A, a<b. Here is where I get stucked. I cannot come up with an idea of an a larger than c but smaller than b.

    Any hint in the right direction will be very much appreciated. Thank you !
     
  2. jcsd
  3. Oct 29, 2012 #2
    How about a proof by contradiction? If for c≥a for all points a in A, what does that tell you about the least upper bound?
     
  4. Oct 29, 2012 #3
    That would mean b is not the least upper bound since c is smaller than b and greater or equal to any a in A which is a contradiction since by hypothesis b is the supremum of A.

    Awesome !
    Thank you clamtrox !
     
  5. Oct 30, 2012 #4


    Now I am trying to prove the statement in the other direction:
    Let a,b,c be reals and let A be a set of real numbers. If c<b and there is an a in A such that a>c then b is the supremum of A.

    From the givens I know that:
    c<b
    There is an a in A such that a>c.

    What I need to prove is that for any a in A, a<=b.
    I took a>c from the givens but that is where I get stucked because I do not know how to show that this a is greater or equal to b.

    Do you have any hint?
    Thank you in advance !
     
  6. Oct 30, 2012 #5
    There's something wrong here... Let a=1, A={1}, c=0 and b=42. Then c<b and a>c exists, yet b is not the supremum. Maybe it should say "If for all c<b there exists a in A s.t. a>c ... "
     
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